Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int n;
int main() {
  cin>>n;
  double x;
  double dp[n+1];
  for(int i=0;i<=n;i++)
    dp[i] = 0;
  dp[0]=1;
  for(int i=1;i<=n;i++)
  {
    cin>>x;
    for(int j=i;j>0;j--)
    {
      dp[j] = x*dp[j-1] + (1-x)*dp[j]; 
    }
    dp[0] = (1-x)*dp[0];
  }
  double ans=0;
  for(int i=n/2+1;i<=n;i++)
  {
    ans+=dp[i];
  }
  cout<<fixed<<setprecision(10)<<ans;
  return 0;
}